Submitted by,

Registration no- 2013334038 2013334039 2013334040 2013334041

Submitted to,Mr. Chowdhuury Md Lutfur Rahman

Assistant Professor,Department of Industrial and Production Engineering, Shahjalal University of Science and Technology.

PRSENTATION ON ONE DIMENSIONAL STEADY STATE FIN EQUATION (LONG FIN)

ONE DIMENSIONAL STEADY STATEFIN EQUATION FOR LONG FIN

FINSWhen heat transfer by convection between a surface and the fluid surrounding it can be increased by attaching to the surface thin strips of metal called fins.

Figure: Fin

In heat change applications a finned tube arrangement might be used to remove heat from a hot liquid. The heat transfer from the liquid to the finned tube is by convection. The heat is conducted through the material and finally dissipated to the surrounding by convection.

ONE DIMENSIONAL STEADY STATE FIN EQUATION (LONG FIN)

Figure: Nomenclature for the derivation of one dimensional fin equation.

Consider one dimensional fin exposed to a surrounding fluid at temperature T. The temperature at the base (x=0) of fin . To develop a one dimensional steady state equation for fin of uniform cross- section. Let us consider a differential volume element of thickness dx. The surface area for convection, of which is pdx∕hpdx(T-T∞).

ONE DIMENSIONAL STEADY STATE FIN EQUATION (LONG FIN)

Figure: Nomenclature for the derivation of one dimensional fin equation.

The steady state energy balance equation for this volume element is:

qx qx+dx + dqconvec

qx

qx+dx

qconvec

; When,

Energy in left face = Energy out in right face + Energy lost by convection

ONE DIMENSIONAL STEADY STATE FIN EQUATION (LONG FIN)

Figure: Nomenclature for the derivation of one dimensional fin equation.

And m=

Which is one dimension linear homogeneous equation.

AT x=0, At ,

For homogeneous equation the solution is:

1e-mx +C2 emx

= T(x) –T(∞)

ONE DIMENSIONAL STEADY STATE FIN EQUATION (LONG FIN)

Figure: Nomenclature for the derivation of one dimensional fin equation.

At x = 0, = C1 At , C2=0

θ(x)= θ˳e-mx

=e-mx

=e-mx

This is the temperature distribution equation for long fin.

Heat FlowQ = -KA

= KAm =

This is the heat flow equation for long fin.

Problem A steel rod diameter D=2cm, length L=25 Cm and thermal conductivity K=50w∕m˳c is exposed to ambient air at T=20°c

with a heat transfer coefficient h=64w∕m2ᵒ-c. If one end of the rod is maintained at temperature of 120ᵒ.calculate heat loss from the rod.

m2 ====

m=16 mL=16

q==

=(120-20)

=25.1 w (Answer).

Solution

Exercise Consider a long, slender copper rod of diameter D=1 cm and thermal conductivity K=380W/(mᵒc), with one end

thermally attached to a wall at 200ᵒc.Heat is dissipated from the rod by convection with a heat transfer coefficient h∞=15W/ . Determine the heat transfer rate from the rod Into the surrounding air at T∞=30.

m2 ==== m = 3.98 ML=3.98 q== =(200-30) = 20.16W (Answer).

Solution